29_ch 03 Mechanical Design budynas_SM_ch03

29_ch 03 Mechanical Design budynas_SM_ch03 - x = ( l + a )...

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42 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-27 (a) Model (c) I = π 64 (0 . 5 4 ) = 3 . 068(10 3 )in 4 A = π 4 (0 . 5 2 ) = 0 . 1963 in 2 σ = Mc I = 218 . 75(0 . 25) 3 . 068(10 3 ) = 17 825 psi = 17 . 8 kpsi Ans. τ max = 4 3 V A = 4 3 500 0 . 1963 = 3400 psi Ans. (b) Model (d) M max = 500(0 . 25) + 1 2 (500)(0 . 375) = 218 . 75 lbf · in V max = 500 lbf Same M and V σ = 17.8 kpsi Ans. τ max = 3400 psi Ans. 3-28 q =− F ± x ² 1 + p 1 ± x l ² 0 p 1 + p 2 a ± x l ² 1 + terms for x > l + a V =− F + p 1 ± x l ² 1 p 1 + p 2 2 a ± x l ² 2 + terms for x > l + a M =− Fx + p 1 2 ± x l ² 2 p 1 + p 2 6 a ± x l ² 3 + terms for x > l + a At
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Unformatted text preview: x = ( l + a ) + , V = M = 0, terms for x > l + a = F + p 1 a p 1 + p 2 2 a a 2 = p 1 p 2 = 2 F a (1) l p 2 p 1 a b F 1.25" 500 lbf 500 lbf 0.25" 1333 lbf/in 500 500 O V (lbf) O M M max 1.25 in 500 lbf 500 lbf 500 lbf 500 lbf 0.4375 500 500 O V (lbf) O M (lbf in) M max 500(0.4375) 218.75 lbf in...
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This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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